The pole of the straight line `9x+ y - 28=0` with respect to the circle `x^(2)+y^(2)-6x -8y +5=0` is

Options

A. (3, 1)
B. (1,3 )
C. (3, -1)
D. none

Answer

To find the pole of the straight line \(9x + y - 28 = 0\) with respect to the circle \(x^2 + y^2 - 6x - 8y + 5 = 0\), we follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 6x - 8y + 5 = 0 \]
We can complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 6x \quad \text{can be rewritten as} \quad (x - 3)^2 - 9 \]

For \(y\):
\[ y^2 - 8y \quad \text{can be rewritten as} \quad (y - 4)^2 - 16 \]

Substituting these back into the circle equation:
\[ (x - 3)^2 - 9 + (y - 4)^2 - 16 + 5 = 0 \]
This simplifies to:
\[ (x - 3)^2 + (y - 4)^2 - 20 = 0 \]
Thus, the equation of the circle in standard form is:
\[ (x - 3)^2 + (y - 4)^2 = 20 \]
This shows that the center of the circle is at \((3, 4)\) and the radius is \(\sqrt{20}\).

### Step 2: Identify the Coefficients
The coefficients of the line \(9x + y - 28 = 0\) are:
- \(A = 9\)
- \(B = 1\)
- \(C = -28\)

### Step 3: Use the Formula for the Pole
The pole of the line \(Ax + By + C = 0\) with respect to the circle \((x - h)^2 + (y - k)^2 = r^2\) can be found using the formulas:
\[ h' = \frac{-A}{2} + h \]
\[ k' = \frac{-B}{2} + k \]
Where \((h, k)\) is the center of the circle.

Substituting the values:
- \(h = 3\)
- \(k = 4\)

Calculating \(h'\) and \(k'\):
\[ h' = \frac{-9}{2} + 3 = -\frac{9}{2} + \frac{6}{2} = -\frac{3}{2} \]
\[ k' = \frac{-1}{2} + 4 = -\frac{1}{2} + \frac{8}{2} = \frac{7}{2} \]

### Step 4: Conclusion
Thus, the pole of the line \(9x + y - 28 = 0\) with respect to the circle is:
\[ \left(-\frac{3}{2}, \frac{7}{2}\right) \]